Answer
Choice $B$
Work Step by Step
Dividing each term of the given polynomial by the given monomial, then
$$\begin{aligned}
&(2x^2+4x+2)\div(2x)
\\&=
\frac{2x^2}{2x}+\frac{4x}{2x}+\frac{2}{2x}
\\&=
x+2+\frac{2}{2x}
.\end{aligned}
$$
The long division method above shows that the remainder is $2$, which is not negative. Therefore, the given Statement I is incorrect.
The dividend, $2x^2+4x+2$, is in standard form since the terms are arranged in decreasing order of exponents for the variable. Therefore, the given Statement II is correct.
For positive values of $x$, such as $x=3$, the quotient, $x+2$, is equal to
$$
x+2=3+2=5
,$$
while the divisor, $2x$, is equal to
$$
2x=2(3)=6
.$$Since the value of the quotient (i.e. $5$) is less than the value of the divisor (i.e. $6$), then the given Statement III is incorrect.
Since only Statement II is correct, then Choice $B$ is the only true statement.
